php - 使用PHP和MySQL无法获取文件名

Question

我需要帮助。我无法将文件上传到目录。我在下面解释我的代码。

 <?php
    $dbobj = new DBConnection();
    $imageName='newsimage';
    $imagePath="../uploads/news/";
    $dbobj->uploadImage($_FILES,$imageName,$imagePath,function($image){
                print_r($image);exit;
    });

    class DBConnection{
      public function uploadImage($files,$imageFieldName,$imageDirPath,$callback){
       // echo $imageField;exit;
       // print_r($files[$imageFieldName]);exit;
        $result=array();
        $imageName=generateRandomNumber().'_'.$_FILES[$imageFieldName]['name'];
        $target_dir = $imageDirPath;
        $target_file = $target_dir . basename($imageName);
        $uploadOk = 1;
        //echo $files[$imageFieldName]["tmp_name"] ;exit;
        $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
        //echo $imageName;exit;
        if (file_exists($target_file)) {
            $result['msg']="Sorry, file already exists.";
            $result['num']=0;
            $callback($result);
            $uploadOk = 0;
        }
        if ($_FILES[$imageFieldName]["size"] > 500000) {
            $result['msg']="Sorry, file size is large.";
            $result['num']=0;
            $callback($result);
            $uploadOk = 0;
        }
        if($imageFileType != "jpg") {
            $result['msg']="Sorry, only .jpg,.jpeg and .png files are allowed.";
            $result['num']=0;
            $callback($result);
            $uploadOk = 0;
        }
        if ($uploadOk == 0) {
            $result['msg']="Sorry, Your file could not uploaded.";
            $result['num']=0;
            $callback($result);
        }else{
            if (move_uploaded_file($_FILES[$imageFieldName]['tmp_name'], $target_file)) {
                $result['msg']="Image has uploaded successfully.";
                $result['num']=1;
                $callback($result);
            }else{
                $result['msg']="Sorry, Your Image could not uploaded to the directory.";
                $result['num']=0;
                $callback($result);
            }
        }
   }
    }
    function generateRandomNumber(){
        $result=base_convert((float)rand()/(float)getrandmax() * round(microtime(true) * 1000), 10, 36);
        return $result;
    }
    ?>

在这里，我试图将 $_FILES 设置为参数。在 uploadImage 函数中，我试图回显图像名称，但我没有得到它。所以我不能将文件上传到文件夹中。请帮我解决此问题。

Answer 1

$_POST、$_GET 和 $_FILES 是 super 全局变量，它们始终可用。因此将它们作为函数参数传递是个坏主意。

而不是将 $_FILES 作为函数参数传递，尝试在函数本身中获取它。

class DBConnection{
  public function uploadImage($imageFieldName,$imageDirPath,$callback)
  {
     echo $_FILES[$imageFieldName]['name'];
     echo $_FILES[$imageFieldName]['tmp_name'];
  }

}

希望对你有帮助。